Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, z\neq 0$. $\dfrac{{(r^{3})^{-4}}}{{(r^{4}z)^{-3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{3}}$ to the exponent ${-4}$ . Now ${3 \times -4 = -12}$ , so ${(r^{3})^{-4} = r^{-12}}$ In the denominator, we can use the distributive property of exponents. ${(r^{4}z)^{-3} = (r^{4})^{-3}(z)^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{3})^{-4}}}{{(r^{4}z)^{-3}}} = \dfrac{{r^{-12}}}{{r^{-12}z^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-12}}}{{r^{-12}z^{-3}}} = \dfrac{{r^{-12}}}{{r^{-12}}} \cdot \dfrac{{1}}{{z^{-3}}} = r^{{-12} - {(-12)}} \cdot z^{- {(-3)}} = z^{3}$.